All About Circuits site states that waveforms that are symmetrical above and below their horizontal centerlines contain no even-numbered harmonics. Can somebody explain this mathematically, or point to a resource? I do not know much about dsp, but understand what a Fourier transform is.
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$\begingroup$ 0 Hz is an even (0th) harmonic and if you have a constant bias you will have it. To close this loophole, "horizontal centerline" can be replaced by "zero level". $\endgroup$ – Olli Niemitalo Feb 11 at 6:37
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The complex Fourier coefficients of a $T$-periodic function $f(t)$ are given by
$$c_n=\frac{1}{T}\int_{0}^Tf(t)e^{-j2\pi nt/T}dt\tag{1}$$
with
$$f(t)=\sum_{n=-\infty}^{\infty}c_ne^{j2\pi nt/T}\tag{2}$$
The coefficients with even indices are
$$\begin{align}c_{2n}&=\frac{1}{T}\int_{0}^Tf(t)e^{-j4\pi nt/T}dt\\&=\frac{1}{T}\left[\int_{0}^{T/2}f(t)e^{-j4\pi nt/T}dt+\int_{T/2}^{T}f(t)e^{-j4\pi nt/T}dt\right]\\&=\frac{1}{T}\left[\int_{0}^{T/2}f(t)e^{-j4\pi nt/T}dt+\int_{0}^{T/2}f(t+T/2)e^{-j4\pi nt/T}e^{-j2\pi n}dt\right]\\&=\frac{1}{T}\int_{0}^{T/2}\left[f(t)+f(t+T/2)\right]e^{-j4\pi nt/T}dt\\&=\frac12\frac{2}{T}\int_{0}^{T/2}\left[f(t)+f(t+T/2)\right]e^{-j2\pi nt/(T/2)}dt\\&=\frac12 d_n\tag{3}\end{align}$$
where $d_n$ are the complex Fourier coefficients of the $T/2$-periodic function $g(t)=f(t)+f(t+T/2)$. These coefficients can only be zero if $g(t)=0$, i.e., if $f(t)=-f(t+T/2)$. This latter condition is exactly the symmetry condition you mention in your question. Consequently, the even Fourier coefficients are zero if (and only if) $f(t)$ satisfies
$$f(t)=-f(t+T/2)\tag{4}$$
Or, in other words, a $T$-periodic function $f(t)$ has only odd harmonics if it is $T/2$-antiperiodic. I.e., if you shift the function by half its period and flip it across the horizontal axis, it must look the same as before.
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An $M\text{-}$periodic waveform that has half-periods of length $M/2$ that are of opposite signs but of equal absolute value are by definition $M/2\text{-}$Bloch-periodic with a coefficient of $-1:$ Each period of length $M/2$ is identical to the previous period multiplied by the coefficient.
A generalized frequency-shifted discrete Fourier transform (DFT) of length $M/2$ with a frequency shift of $0.5$ bin widths, or $\frac{0.5\times2\pi}{M/2}$ in angular frequency, has extended basis functions that are $M/2\text{-}$Bloch-periodic with a coefficient of $-1.$ The $M/2\text{-}$Bloch-periodic waveform can be represented by a weighted sum of the extended basis functions. The basis functions have angular frequencies $\frac{0.5\times2\pi}{M/2}, \frac{1.5\times2\pi}{M/2}, \frac{2.5\times2\pi}{M/2}, \ldots.$ This can be rewritten as $\frac{1\times2\pi}{M}, \frac{3\times2\pi}{M}, \frac{5\times2\pi}{M}, \ldots.$ The extended basis functions do not contain even harmonics of the $M\text{-}$periodic angular frequency $\frac{2\pi}{M}.$ This property is retained by the waveform which is their weighted sum.
A function that is $N\text{-}$Bloch-periodic with coefficient $-1$ is also called an $N\text{-}$antiperiodic function.
